第 7 章偏心受力构件091210(1).ppt

第,7,章 偏心构件承载力,N,e,0,M=Ne,0,N,N,偏心受力构件,P184,图,7-4,大偏压破坏,cu,N,f,y,A,s,f,y,A,s,N,N,(,a,),(,b,),e,P184,图,7-5,小偏压破坏,N,f,y,A,s,f,y,A,s,N,N,N,s,A,s,s,A,s,cmax2,cmax1,cu,(,a,),(,c,),(,b,),e,e,P184,图,7-5,条件,应力状态,破坏特征,破坏性质,类似构件,大偏压,（受拉）,小偏压,（受压）,偏心受压构件破坏形态,条件,应力状态,破坏特征,破坏性质,类似构件,大偏压,（,受拉,）,小偏压,（,受压,）,偏心受压构件破坏形态,偏心距较大，远侧,As,较,少,部分受拉,部分受压,砼：,fc,远侧,As,：,fy,近侧,As,：,fy,延性,双筋,适筋,梁,偏心距较大，远侧,As,较,多,部分受拉,部分受压,砼：,fc,远侧,As,：,s,s,近侧,As,：,fy,脆性,双筋,超筋,梁,偏心距较小,全截面受压,砼：,fc,远侧,As,：,s,s,近侧,As,：,fy,脆性,轴压构件,b,c,d,e,f,g,h,A,s,A,s,h,0,x,0,x,0b,s,0.0033,a,a,a,y,0.002,P185,图,7-6,偏心受压构件,N-M,相关曲线,M,u,N,u,轴压破坏,弯曲破坏,界限破坏,小偏压破坏,大偏压破坏,2,1,3,N,相同,M,越大越不安全,M,相同：大偏压，,N,越小越不安全,小偏压，,N,越大越不安全,P187,图,7-7,A,B,C,偏心距,N,N,e,i,a,f,e,i,P186,图,7-8,N,0,N,1,N,2,N,0,e,i,N,1,e,i,N,2,e,i,N,1,a,f1,N,2,a,f2,B,C,A,D,E,短柱(材料破坏),长柱(材料破坏),细长柱(失稳破坏),N,M,0,P187,图,7-9,e,f,y,A,s,e,i,a,1,f,c,e,A,s,f,y,N,b,A,s,A,s,a,s,a,s,h,0,h,x,大偏心受压构件承载力基本公式,保证混凝土受压破坏发生在钢筋屈服之后,保证纵向受压钢筋在破坏时达到屈服,P184,图,7-5(a),e,f,y,A,s,e,i,b,a,1,f,c,e,A,s,s,A,s,A,s,a,s,h,N,h,0,x,a,s,小偏心受压构件承载力基本公式,P184,图,7-5(b),两种偏心受压情况的判别,基本条件判别：,大偏心受压构件,小偏心受压构件,进行截面配筋设计时的,初步判别,：,小偏心受压构件,大偏心受压构件,不对称配筋时（,A,s,A,s,）的截面设计,-,大偏压,情形,I,：,A,s,和,A,s,均不知,设计的基本原则：,A,s,+,A,s,为最小,充分发挥混凝土的作用,情形,II,：已知,A,s,求,A,s,2,a,s,2,a,s,另一平衡方程求,A,s,C,e,N,f,y,A,s,f,y,A,s,e,e,i,x,1,f,c,例,1,：偏压柱，,bxh=350mmx500mm,，,l,0,=4.2m,，,N=1200kN,，,M=250kNm,，,C30,砼，,HRB400,级钢筋。求,As,，,As,解（,1,）判别大小偏压,h,e,i,先按大偏压计算,例,1,：偏压柱，,bxh=350mmx500mm,，,l,0,=4.2m,，,N=1200kN,，,M=250kNm,，,C30,砼，,HRB400,级钢筋。求,As,，,As,解（,2,）计算配筋,选配,2 22+1 18,（,As=1015mm,2,）,选配,2 22+1 18,（,As=1015mm,2,）,例,1,：偏压柱，,bxh=350mmx500mm,，,l,0,=4.2m,，,N=1200kN,，,M=250kNm,，,C30,砼，,HRB400,级钢筋。求,As,，,As,解（,3,）复核,x,（,4,）平面外承压验算,解（,1,）判别大小偏压,h,e,i,（,2,）计算配筋,选配,2 20+1 18,（,As=883mm,2,）,例,2,：偏压柱，,bxh=350mmx500mm,，,l,0,=4.2m,，,N=1200kN,，,M=250kNm,，,C30,砼，,HRB400,级钢筋。在受压区配置了,3,根,HRB400,级直径为,22,的钢筋（,As=1140mm2,），求解,As,解（,3,）复核,x,（,4,）平面外承压验算,例,2,：偏压柱，,bxh=350mmx500mm,，,l,0,=4.2m,，,N=1200kN,，,M=250kNm,，,C30,砼，,HRB400,级钢筋。在受压区配置了,3,根,HRB400,级直径为,22,的钢筋（,As=1140mm2,），求解,As,f,y,A,s,f,y,A,s,a,s,a,1,f,c,bx,h,0,a,s,h,h(,e,i,-e,0,),e,N,a,s,小偏心：反向破坏,P191,图,7-11,不对称配筋时（,A,s,A,s,）的截面设计,-,小偏压,设计的基本原则,：,A,s,+,A,s,为最小,C,s,A,s,N,e,e,f,y,A,s,e,i,x,1,f,c,h,0,f,y,A,s,N,e,e,i,f,y,A,s,1,f,c,a,s,几何中心轴,实际力线,e,a,反向破坏,构造要求：,不对称配筋时（,A,s,A,s,）的截面复核,已知,e,0,求轴向力,N,大偏心受压,按大偏心受压计算,联立求解,按小偏心受压计算,大偏心受压,已知,N,求弯矩,M,界限轴向力,大偏心受压,求解,求解,小偏心受压,例,3,：偏压柱，,bxh=300mmx400mm,，,l,0,=3.2m,，,N=300kN,，,C20,砼，配有,2 18+2 22,的受拉钢筋（,As=1269mm,2,），,3 20,的受压钢筋（,As=942mm,2,），求截面在,h,方向能承受的弯矩设计值,M,。,解（,1,）判别大小偏压,属于大偏压,（,2,）求解,x,，,e,例,3,：偏压柱，,bxh=300mmx400mm,，,l,0,=3.2m,，,N=300kN,，,C20,砼，配有,2 18+2 22,的受拉钢筋（,As=1269mm,2,），,3 20,的受压钢筋（,As=942mm,2,），求截面在,h,方向能承受的弯矩设计值,M,。,解（,3,）求解荷载偏心距,e,0,例,3,：偏压柱，,bxh=300mmx400mm,，,l,0,=3.2m,，,N=300kN,，,C20,砼，配有,2 18+2 22,的受拉钢筋（,As=1269mm,2,），,3 20,的受压钢筋（,As=942mm,2,），求截面在,h,方向能承受的弯矩设计值,M,。,解（,4,）求解,M,（,5,）平面外承压验算,例,4,：偏压柱，,bxh=300mmx500mm,，,l,0,=6m,，,C30,砼，配有,2 20,的受拉钢筋（,As=628mm,2,），,3 20,的受压钢筋（,As=942mm,2,），已知,e,0,=80mm,，求截面能承受的轴力设计值,N,。,解（,1,）判别大小偏压,属于小偏压,例,4,：偏压柱，,bxh=300mmx500mm,，,l,0,=6m,，,C30,砼，配有,2 20,的受拉钢筋（,As=628mm,2,），,3 20,的受压钢筋（,As=942mm,2,），已知,e,0,=80mm,，求截面能承受的轴力设计值,N,。,解（,2,）计算,x,，,N,，并判别,x,，,N,将已知数据带入小偏压计算公式,N,即为所求,（,3,）平面外承压验算,对称配筋时（,A,s,A,s,）两种偏心受压构件的判别,界限轴向力,小偏心受压构件,大偏心受压构件,且,且,教材,【,例,7-1】P194,非对称配筋：大偏压截面设计（,As,，,A,s,均未知）,【,例,7-2】P196,非对称配筋：大偏压截面复核（已知,N,，求,M,）,【,例,7-3】P198,对称配筋：大偏压截面设计（,x2as,）,课堂,【,例,1】,非对称配筋：大偏压截面设计（,As,，,A,s,均未知）,【,例,2】,非对称配筋：大偏压截面,设计（,A,s,已知，求,As,）,【,例,3】,非对称配筋：大偏压截面复核,（已知,N,，求,M,）,【,例,4】,非对称配筋：小偏压截面,复核（已知,e,0,，求,N,）,习题,7-6,：偏压柱，,bxh=300mmx400mm,，,l,0,=3m,，,N=300kN,，,M=150kNm,，,C20,砼，,HRB335,级钢筋。采用对称配筋，求,As=As=,？,解（,1,）判别大小偏压,属于大偏压构件,习题,7-6,：偏压柱，,bxh=300mmx400mm,，,l,0,=3m,，,N=300kN,，,M=150kNm,，,C20,砼，,HRB335,级钢筋。采用对称配筋，求,As=As=,？,解（,2,）配筋计算,配有,4 20,（,As=As=1256mm,2,）,（,3,）平面外承压验算,a,1,f,c,f,y,A,s,A,s,A,s,s,s,A,s,f,y,A,s,A,s,A,s,f,y,A,s,N,N,e,e,e,0,e,e,0,e,a,s,a,s,a,s,a,s,h,/2,h,/2,h,/2,h,/2,(,a,),(,b,),和偏压不同,